This question already has an answer here:

Confusion about a question, I am probably missing something, i.e. I'm not sure what the point of the question is.

Let $f_n$ be sequence of $\mu$ almost everywhere finite $\mathbb R$ valued functions, where $\mu$ is a $\sigma$-finite measure. Prove that there exists constants $c_n>0$ such that $\sum_n c_n f_n$ converges for almost every $x$.

Now, because the functions are almost everywhere finite, the only way this will converge is if the $c_n$'s tend to zero as $n$ goes to infinity. I'm confused how to use the measure here, should I be looking at subsets of $X$ and see if the result holds there?


marked as duplicate by user147263, PhoemueX, Alex M., zhoraster, Tom-Tom Oct 14 '15 at 7:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 6
    $\begingroup$ Please, take a look at: math.stackexchange.com/questions/209668/… . $\endgroup$ – Ramiro Oct 14 '15 at 3:25
  • 1
    $\begingroup$ Please, take also a look at: math.stackexchange.com/questions/227228/… . $\endgroup$ – Ramiro Oct 14 '15 at 3:29
  • 1
    $\begingroup$ If the domain of the function $f_n$ is an arbitrary $\sigma$-finite space $X$, it makes NO SENSE to write $x\in[-n,n]$. In this general case, we must use that $X$ is $\sigma$-finite and so $X=\bigcup_{n \in \mathbb{N}} E_n$ where, for each $n$, $\mu(E_n)<+\infty$ and $E_n \subseteq E_n+1$. Then, we define $$ A_n=\{x\in E_n \,\mid\,c_n\cdot|f_n(x)|\gt1/n^2\} $$ Now, we can prove that $c_n$ can be chosen so that $\mu(A_n)<\frac{1}{n^2}$. $\endgroup$ – Ramiro Oct 14 '15 at 13:27
  • 1
    $\begingroup$ Here is the proof that $c_n$ can be chosen so that $\mu(A_n)<\frac{1}{n^2}$. For a FIXED $n$, define $$B_k=\left \{x\in E_n \,\mid\, \frac{1}{k}\cdot|f_n(x)|\gt1/n^2 \right \}$$ where $k\in\mathbb{N}$ and $k\geq1$. It is easy to see that, for every $k$, $B_k\subseteq E_n$ and $B_{k+1}\subseteq B_{k}$ and $\bigcap_{k=1}^\infty B_k=\emptyset$. Since $\mu(E_n)<+\infty$, we have $$ \lim_{k \to \infty} \mu(B_k)=\mu \left ( \bigcap_{k=1}^\infty B_k \right ) = 0 $$ So there is $k_n$ (in fact, many) such that $\mu(B_{k_n})< \frac{1}{n^2}$. Make $c_n=\frac{1}{k_n}$. $\endgroup$ – Ramiro Oct 14 '15 at 13:51
  • 1
    $\begingroup$ Returning to the main proof. We now have $$A_n=\{x\in E_n \,\mid\,c_n\cdot|f_n(x)|\gt1/n^2\} $$ where $c_n$ were chosen so that $\mu(A_n)<\frac{1}{n^2}$. Then, apply Borel-Cantelli lemma we have that $\mu(\limsup A_n)=0$, hence, for $\mu$-almost every $x$ in $X$, $x$ is not in $A_n$ for every $n$ large enough. Since, given any $x\in X$, $x \in E_n$ for every $n$ large enough, this means that $c_n\cdot|f_n(x)|\leqslant1/n^2$ for every $n$ large enough. In particular, the series $\sum\limits_nc_nf_n(x)$ converges (absolutely), for $\mu$-almost every $x$ in $X$. $\endgroup$ – Ramiro Oct 14 '15 at 14:06

Browse other questions tagged or ask your own question.